A bag contains 3 blue and 5 white marbles. One by one, marbles are drawn out randomly until only two are left in the bag. What is the probability that out of the two, one is white and one is blue?

A. 15/56
B. 41/56
C. 13/28
D. 15/28
E. 5/14

Good question. +1

The required probability = probability of choosing 6 balls out of the total 8 in such a way that we remove 4 out of 5 white and 2 out of 3 blue balls.

Ways to select 6 out of total 8 = 8C6

Ways to select 4 out of 5 white balls = 5C4

Ways to select 2 out of 3 blue balls = 3C2

Thus the required probability = (5C4*3C2)/8C6 = 15/28.

D is thus the correct answer.

Hello VeritasPrepKarishma, can you explain how this is the same.

Honestly the easiest way to solve this is to draw it out and eyeball it.

B B B W W W W W

Visually, you can see that there are 15 combinations of one blue marble and one white marble (the first "B" combined with all 5 "W" + the second "B" combined with all 5 "W" + the third "B" combined with all 5 "W" = 15).

Then just use the combination formula for 8C2 to find the total number of combinations.

8!/2!*6! = 8*7/2 = 28

Answer is thus 15/28

Let us try to make this simpler. If the question had instead asked:

"A bag contains 3 blue and 5 white marbles. One by one, marbles are drawn out randomly until only ONE is left in the bag. What is the probability that it is blue?"

We can say that there are two equivalent ways of going about this:
1. We take out 7 marbles and leave one in the bag.
2. We take out one and leave 7 in the bag.

These two operations are equivalent. After all, we are randomly separating one marble from the other seven. Hence, these two experiments will yield the same probability.

You can now extend this to the case where 2 marbles need to be separated from the rest.

oh damn...can't believe I missed this one...
overcomplicated myself...
8C6=28
5C4 * 3C2 = 5x3 = 15
15/28

damn...

Another quick solution:

Prob. that 1B & 1W remains in the bag = (Number of ways we can select 6 balls -2B & 4W - from the Bag) / (Total number of ways we can choose 6 balls from the bag)

Numerator = 6P6 / (2!*4!) = 15
Denominator = 8C6 = 28
Required prob. = 15/28

Thanks.

We need to determine the probability of selecting 2 blue marbles and 4 white marbles in 6 selections. The number of ways to select the blue marbles is 3C2 = 3 and the number of ways to select the white marbles is 5C4 = 5. Thus, the total number of ways is 3 x 5 = 15.

The total number of ways to select 6 marbles from 8 is 8C6 = (8 x 7 x 6 x 5 x 4 x 3)/6! = (8 x 7 x 6 x 5 x 4 x 3)/(6 x 5 x 4 x 3 x 2) = (8 x 7)/2 = 28.

Thus, the probability is 15/28.

Answer: D
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Another way to think of it...

If you arrange the 3 blue and 5 white in a row, how many of those arrangements end in BLUE-WHITE or WHITE-BLUE?

Total # of arrangements of 3 blue, 5 white: 8!/3!5! = 56
# of arrangements ending in B-W or W-B = (arrange 2 blue and 4 white) * 2 = (6!/2!4!) * 2 = 30

So, 30/56 (15/28) ways to arrange 3 blue and 5 white in a row such that the last 2 marbles are B-W or W-B.

Answer: D